B U (A C): Fill & Download for Free

I-S T-H-I-S E-N-O-U-G-H T-O C-O-M-M-U-N-I-C-A-T-E W-I-T-H Y-O-U? I A-M Q-U-I-T-E S-U-R-E T-H-A-T B-Y N-O-W Y-O-U A-R-E F-I-N-D-I-N-G T-H-I-S S-O-M-E-W-H-A-T A-N-N-O-Y-I-N-G. S-U-R-E, I-T I-S B-E-T-T-E-R T-H-A-N N-O-T-H-I-N-G A-N-D I-T W-I-L-L D-O O-N A S-H-O-R-T T-E-R-M B-A-S-I-S F-O-R S-H-O-R-T M-E-S-S-A-G-E-S B-U-T I-T W-I-L-L N-E-V-E-R B-E A-S S-A-T-I-S-F-A-C-T-O-R-Y A-S C-O-M-M-U-N-I-C-A-T-I-N-G I-N A N-A-T-U-R-A-L M-A-N-N-E-R S-U-C-H A-S I-N A N-A-T-U-R-A-L S-I-G-N-E-D L-A-N-G-U-A-G-E.

The new quadratic formula, from Dr. Po Shen Loh of CMU, is essentially[math]x^2 - 2bx + [/math][math]c[/math][math] = 0[/math] has roots [math]x = [/math][math]b[/math][math] \pm \sqrt{b^2 - [/math][math]c[/math][math]}[/math]It’s not really that new; I’ve been hyping it on Quora as The Shakespeare Quadratic Formula (2b or -2b) since 2016. Dean Rubine's answer to How would I solve for x in formula: x + \dfrac{1}{x} = 8?I’d say Dr. Loh’s derivation is novel, or as novel as can be on this well trodden territory. I’ll summarize it as:To factor [math]x^2 - 2bx + [/math][math]c[/math][math] = 0[/math] we need two numbers which add to [math]-2b[/math] and multiply to [math]c[/math][math].[/math] Since they add to [math]-2b[/math] they average to [math]-b.[/math] So they’re on opposite sides of [math]-b[/math], equal distance away, let’s call them [math]-b-u[/math] and [math]-b+u[/math] for some [math]u[/math][math]. [/math] So[math]c[/math][math] = (-b-u)(-b+u) = b^2 - u^2[/math][math]u^2 = b^2 -c [/math][math]u[/math][math] = \pm \sqrt{b^2 -c}[/math]So our quadratic factors as[math](x + (-b - \sqrt{b^2-c})))(x + (-b + \sqrt{b^2-c}))= 0[/math]so roots[math]x = [/math][math]b[/math][math] \pm \sqrt{b^2-c}[/math]Dr. Loh’s procedure is more a method for factoring a given quadratic equation than a new quadratic formula. He never writes the general quadratic with a coefficient on a coefficient ([math]-2[/math] on the [math]b[/math][math][/math] above); that’s my explanation. I’ve only seen him solve particular quadratics using his method.I find the Loh method to be slightly indirect. We’re after the roots of the quadratic; the factors are a passing step. A quadratic equation [math]x^2-Bx+C=0[/math] is the mathematical equation to find two numbers which add to [math]B[/math][math][/math] and multiply to [math]C[/math][math].[/math] So we could apply Dr. Loh’s method to the roots rather than the factors, which just amounts to using [math]b\pm [/math][math]u[/math][math][/math] instead of [math]-b \pm [/math][math]u[/math][math][/math] above, and skip the factoring to get the roots directly.

This looks tedious to do manually, so I'll write a Haskell program to do it.First, let's define a data type for a set-arithmetic expression:data Expr = Union Expr Expr  | Intersect Expr Expr  | Complement Expr  | Variable String  deriving (Show) Now to prove the identity, we need to check that for any possible element in the universe, it's either in the set on both sides of the equations, or it isn't. To check the equation for a particular element, we "assign" any set (of A, B, C, X, Y) to True if the element is in that set, and False otherwise, evaluate the logical expressions on both sides and see if they come up the same. So lets's write a function which can evaluate an Expr given such an assignment of sets to boolean values.type VariableAssignment = String -> Bool  evalExpr :: VariableAssignment -> Expr -> Bool evalExpr va (Union e1 e2) = (evalExpr va e1) || (evalExpr va e2)  evalExpr va (Intersect e1 e2) = (evalExpr va e1) && (evalExpr va e2)  evalExpr va (Complement e) = not (evalExpr va e) evalExpr va (Variable v) = va v Now we write a function which, given two expressions, will iterate over all [math]2^n[/math] such assignments and check they evaluate the sameequivExprs :: Expr -> Expr -> Bool equivExprs e1 e2 =   let uniqVars = nub (allVars e1 ++ allVars e2)   in all (\trueVars ->  let va = (`elem` trueVars)  in evalExpr va e1 == evalExpr va e2  ) (subsequences uniqVars)   where  allVars :: Expr -> [String]  allVars (Union e1 e2) = allVars e1 ++ allVars e2  allVars (Intersect e1 e2) = allVars e1 ++ allVars e2  allVars (Complement e) = allVars e  allVars (Variable v) = [v]  Now, I don't want to manually type out your expressions into my Expr data type, so I'll just write a quick parser using Parsec:exprParser :: GenParser Char () Expr exprParser = do  lhs <- exprParser1  (do   op <- oneOf "∩U"  rhs <- exprParser  return $ (if op == '∩' then Intersect else Union) lhs rhs   ) <|> (do   return lhs   )   exprParser1 :: GenParser Char () Expr exprParser1 =  (char '(' *> exprParser <* char ')')  <|> (char '[' *> exprParser <* char ']')  <|> ((Variable . (: [])) <$> oneOf acceptableVarNames)  <|> (Complement <$> (char 'U' *> char '\\' *> exprParser1))  where acceptableVarNames = "ABCXY"  parseExpr :: String -> Expr parseExpr s = case parse (exprParser <* eof) "" s  of Left er -> error ("Error parsing: " ++ (show er))  Right res -> res  Finally, let's run this:main = putStrLn $ show $ equivExprs  (parseExpr "[(A∩B)U(A∩C)U((U\\A)∩(U\\X)∩Y)]∩U\\[(A∩(U\\B)∩C)U((U\\A)∩(U\\X)∩(U\\Y))U((U\\A)∩B∩Y)]")  (parseExpr "(A∩B)U((U\\A)∩(U\\B)∩(U\\X)∩Y)") The program prints True. So your identity holds! Voilà.